Question

The energy of an electron in an excited hydrogen atom is $-3.4\text{eV}$. Its angular momentum will be:
$(\text{Take}h=6.626\times {10}^{-34}\text{Js})$

• A. $1.11\times {10}^{34}\text{Js}$
• B. $1.51\times {10}^{-31}\text{Js}$
• C. $2.11\times {10}^{-34}\text{Js}$
• D. $3.76\times {10}^{-34}\text{Js}$

Answer 1

The correct option is C $2.11\times {10}^{-34}\text{Js}$

According to Bohr's atom model, the energy of the stationary states of a hydrogen atom is given by,

$E_n=-\frac{13.6}{n^2}\text{eV}$

As given in the question,

$\Rightarrow -3.4\text{eV}=\frac{-13.6\text{eV}}{n^2}$

$\Rightarrow n^2=4\Rightarrow n=2$

According to Bohr's postulate, Angular momentum of the stationary states is given by,

$L=\frac{nh}{2\pi }=\frac{2h}{2\pi }$

$\Rightarrow L=\frac{h}{\pi }$

$\Rightarrow L=\frac{6.626\times {10}^{-34}}{3.14}=2.11\times {10}^{-34}\text{Js}$

Hence, option $\left(\text{C}\right)$ is correct.

Why this question? Key idea : For Hydrogen atom, energy $E_n=-\frac{13.6}{n^2}\text{eV}$